My Math Forum Graphing a circle TRIGONOMETRY

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 April 10th, 2009, 06:14 PM #1 Senior Member   Joined: Feb 2008 Posts: 100 Thanks: 0 Graphing a circle TRIGONOMETRY Ok I'm really stuck on a problem from my Trig book. Graph the following circle. Question 41. x^2 + y^2 = 25 So first off I had a hard time graphing the circle but on a calculator I just entered y1 = ?5^2 + x^2 and y2 = -?5^2 + x^2. From the equation I figured the radius is 5 and this is how: The equation only has x + y which means the circle is centered on the origin. Now having a triangle inside the circle, x is the base length, y is the height, and 5 is the radius. In other words or in a equation the triangle is x^2 + y^2 = 5^2. *Edited* Actually I'm having a really hard time right now figuring out how I figured out the radius in the first place. Here's the part that's really frustrating me. Question 49. Use the graph of Problem 41 to name the points at which the line x + y = 5 will intersect the circle x^2 + y^2 = 25. In my solutions manual it gave me (0,5) and (5,0) as the point where the line will intersect. I also put that line in the diagram below and I labeled it but photobucket cropped out the label. My question is how would you figure out that line? By solving for y and substituting for x and y? And to graph a circle on graph paper with the above equation how would I do that?
 April 11th, 2009, 12:37 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Graphing a circle TRIGONOMETRY Take a sheet of graph paper, draw the x and y axes, and plot two points at random at the corners of two of the little squares. Convince yourself that if the first point has coordinates (a,b), and the second (c,d), then the distance between them is $\sqrt{(c-a)^2+(d-b)^2}$. (Use Pythagoras's theorem.) Now the definition of a circle is: the set of points a fixed distance away from a central point. The fixed distance is the radius. So if the centre is (a,b), then the equation of the circle will be $\sqrt{(x-a)^2+(y-b)^2}=r$ for a positive radius r. Square both sides to get: $(x-a)^2+(y-b)^2= r^2$ All circles can be written in this form. =============================================== Now, what happens if both a and b are 0? $\\(x-0)^2+(y-0)^2 = r^2 \\x^2+y^2 = r^2$ This is the equation of a circle with centre (0,0). =============================================== To find the intersections of x+y=5 and your circle, the straightforward algebra thing to do is to first rewrite x+y=5 as y=5-x. Then, replace all instances of y in $x^2+y^2= r^2$ with (5-x), and expand. $\\x^2+(5-x)^2 = 25 \\x^2+25-10x+x^2 = 25 \\2x^2-10x=0 \\x(2x-10)=0$ So either x=0 or 2x-10=0. You have two values for x, which you can substitute back into y=5-x to find the corresponding y.
 April 11th, 2009, 12:58 AM #3 Senior Member   Joined: Feb 2008 Posts: 100 Thanks: 0 Re: Graphing a circle TRIGONOMETRY Ok I understand how you get the radius for a circle from this equation $\sqrt{(x-a)^2+(y-b)^2}=r$. Basically there's an unseen triangle with an hypotenuse that is the radius as well. Then the base and height make up for the x and y units, which is a point on a circle, that is if it's origin is on 0,0. If it's not on 0,0 then instead of $\sqrt{(x)^2+(y)^2}=r$ you would have $\sqrt{(x1-x2)^2+(y1-y2)^2}=r$ which also would give you the point on the circle. Correct me if I'm wrong so far. But the part I don't understand is how does this equation $\sqrt{(x-a)^2+(y-b)^2}=r$ make a circle? That might not be the best phrased question but the pieces or puzzle isn't connecting for me.
 April 11th, 2009, 01:04 AM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Graphing a circle TRIGONOMETRY The equation x=5 is used to mean "the set of all points such that the x-coordinate is 5." The equation x+y=5 is used to mean "the set of all points such that the sum of x and y coordinates is 5." So, the equation $\sqrt{(x-a)^2+(y-b)^2}=r$ can denote "the set of all points such that the distance from (a,b) is r."
April 11th, 2009, 04:10 AM   #5
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Quote:
 Originally Posted by David_Lete I'm really stuck on a problem from my Trig book.
You haven't explained precisely why you are stuck.

Quote:
 Originally Posted by David_Lete I had a hard time graphing the circle but on a calculator I just entered y1 = ?5^2 + x^2 and y2 = -?5^2 + x^2.
You won't get a circle from those equations!

Quote:
 Originally Posted by David_Lete Use the graph of Problem 41 to name the points at which the line x + y = 5 will intersect the circle x^2 + y^2 = 25.
x + y = 5 implies x² + 2xy + y² = 25. Subtracting x² + y² = 25 gives 2xy = 0. Hence the two equations are simultaneously satisfied when x = 0 (which implies y = 5) and when y = 0 (which implies x = 5). However, the question didn't ask for an algebraic derivation of these points.

Quote:
 Originally Posted by David_Lete And to graph a circle on graph paper with the above equation how would I do that?
How would you draw any circle of a predetermined size on any piece of paper? Have you considered buying an appropriate drawing instrument?

April 11th, 2009, 04:20 AM   #6
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Quote:
 Originally Posted by David_Lete Graph the following circle. Question 41. x^2 + y^2 = 25
To learn how to graph a circle from its conics-form equation, try here.

Short version: For a circle with equation (x - h)^2 + (y - k)^2 = r^2, draw a dot (for the center) at the point (h, k). Draw four more dots, each "r" units from the center, in the directions up, down, left, and right. Then sketch the circle that passes through these four points.

Quote:
 Originally Posted by David_Lete on a calculator I just entered y1 = ?5^2 + x^2 and y2 = -?5^2 + x^2.
To learn how to solve literal equations, try here. To learn how to work with radical equations, try here

Short version: To isolate y in x^2 + y^2 = 25, you need to start by subtracting the x^2 from either side.

Quote:
 Originally Posted by David_Lete Question 49. Use the graph of Problem 41 to name the points at which the line x + y = 5 will intersect the circle x^2 + y^2 = 25.
To learn how to solve systems of non-linear equations, try here.

Short version: If y = 5 - x, then plug "5 - x" in for "y" in "x^2 + y^2 = 25", giving you the equation x^2 + (5 - x)^2 = 25, which is a quadratic that you can solve by any of various methods, such as the Quadratic Formula.

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